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i. Carbon dioxide (CO2) is a major greenhouse gas when it diffuses into the earth atmosphere. Given that at 273 K and 1 atm, the value of D for CO2 (Dco2) is 1.00 x 10-5 m? s1, calculate the collisional cross-section of the gas. ii. Under the same T and P, the DN2 value of nitrogen gas is three-fold greater than the value of Dco2. Calculate the value of o for N2.

i. Carbon dioxide (CO2) is a major greenhouse gas when it diffuses into the earth atmosphere. Given that at 273 K and 1 atm, the value of D for CO2 (Dco2) is 1.00 x 10-5 m? s1, calculate the collisional cross-section of the gas. ii. Under the same T and P, the DN2 value of nitrogen gas is three-fold greater than the value of Dco2. Calculate the value of o for N2.

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Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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. i. Carbon dioxide (CO2) is a major greenhouse gas when it diffuses into the earth atmosphere. Given that at 273 K and 1 atm, the value of D for CO2 (Dco2) is 1.00 x 10-5 m? s1, calculate the collisional cross-section of the gas. Under the same T and P, the DN2 value of nitrogen gas is three-fold greater than the value of Dco2. Calculate the value of o for N2. ii.
Transcribed Image Text:. i. Carbon dioxide (CO2) is a major greenhouse gas when it diffuses into the earth atmosphere. Given that at 273 K and 1 atm, the value of D for CO2 (Dco2) is 1.00 x 10-5 m? s1, calculate the collisional cross-section of the gas. Under the same T and P, the DN2 value of nitrogen gas is three-fold greater than the value of Dco2. Calculate the value of o for N2. ii.
Solution T= 273K P= 1atm -S Deo = 1.00 x10 m*/s CO %3D DN,= 3 Deg. Now, The diffusion Co efficientgiren by where, A= mean free path mean 1. 3 Vmean = mean Speed. X= KT 8RT %3D man IT M. I KT 8RT TTM 8 x 8.314 X273 3 O cor X 1.01325 X1f Nimz 3.14159 X44x16 -28 123.9378 x1o x 362.4351 mts %3D 24 mi/3 a 4.492 x10 y.492X 10m -59 1.00 X 10 4.492 X104 m²
Transcribed Image Text:Solution T= 273K P= 1atm -S Deo = 1.00 x10 m*/s CO %3D DN,= 3 Deg. Now, The diffusion Co efficientgiren by where, A= mean free path mean 1. 3 Vmean = mean Speed. X= KT 8RT %3D man IT M. I KT 8RT TTM 8 x 8.314 X273 3 O cor X 1.01325 X1f Nimz 3.14159 X44x16 -28 123.9378 x1o x 362.4351 mts %3D 24 mi/3 a 4.492 x10 y.492X 10m -59 1.00 X 10 4.492 X104 m²
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